Question

A 0.5 kg block of ice is sliding by you on a very slippery floor at 4 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.0025 seconds. The eventually slides at an angle of 24 degrees from its original direction labeled θ in the diagram . The overhead view shown in the diagram is approximately to scale. The arrow represents the av force your toe applies briefly to the block of ice Which of the possible paths shown in the diagram corresponds to the correct overhead view of the block's path? Which components of the block's momentum are changed by the impulse applied by your foot? (Check all that apply. The diagram shows a top view, looking down on the xz plane.) x component z component y component What is the unit vector in the direction of the block's momentum after the kick? p-(2,0,.89)x What is the x-component of the block's momentum after the kick? Remember that pPlp. What is the magnitude of the block's momentum after the kick? Use your answers to the preceding questions to find the z-component of the block's momentum after the kick (drawing a diagram is helpful): What was the magnitude of the average force you applied to the block? kg m/s kg m/s kg m/s avgl

Answer 1

As block slides in x-z plane makin angle of 24 deg. with +ve x towards -ve z axis, unit vector in the direction of block's momentum is < cos24 , 0 , - sin24> = < 0.91 , 0, 0.41>

There is no change in x component of momentum.

Hence Pfx = initial momentum in x direction = 0.5*4= 2 kgm/s

Magnitu of momentum, after kick,P= Pfx / cos 24 = 2.2 kgm/s

z component of block's momentum Pz =- P sin24

= - 0.89 kgm/s

Magnitude of average force = Pz / time of contact

= 356 N